Refrigerant-134a enters the coils of the evaporator of a refrigeration system as a saturated liquid-vapor mixture at a pressure of 160 kpa. the refrigerant absorbs 180 kj/kg of heat from the cooled space, which is maintained at -5°c, and leaves as a saturated vapor at the same pressure. determine

(a) the entropy change of the refrigerant,

(b) the entropy change of the cooled space, and

(c) the total entropy change for this process.

(a) 0.699 kJ/K

(b) -0.671 kJ/K

(c) 0.028 kJ/K

Explanation:

The Refrigerant-134a flows into the evaporator as a saturated liquid-vapor mixture and flows out as a saturated vapor at a saturation pressure of 160 kPa and temperature of -15.64°C (estimated from the Saturated Refrigerant-134a Temperature Table).

(a) The entropy change of the refrigerant (ΔS) = Q/T

Q = 180 kJ

T = -15.64 + 273.15 = 257.51 K

ΔS = Q/T = 180/257.51 = 0.699 kJ/K

(b) The entropy change (ΔS) of the cooled space (ΔS) = -Q/T

Q = -180 kJ

T = -5 + 273.15 = 268.15 K

ΔS = Q/T = -180/268.15 = -0.671 kJ/K

(c) The total entropy change for this process (ΔS) = ΔS + ΔS = 0.699 - 0.671 = 0.028 kJ/K

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml